Question: You have found the following ages (in years) of all 5 gorillas at your local zoo: $ 24,\enspace 9,\enspace 27,\enspace 14,\enspace 21$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{24 + 9 + 27 + 14 + 21}{{5}} = {19\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $24$ years $5$ years $25$ years $^2$ $9$ years $-10$ years $100$ years $^2$ $27$ years $8$ years $64$ years $^2$ $14$ years $-5$ years $25$ years $^2$ $21$ years $2$ years $4$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{25} + {100} + {64} + {25} + {4}} {{5}} $ $ {\sigma^2} = \dfrac{{218}}{{5}} = {43.6\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{43.6\text{ years}^2}} = {6.6\text{ years}} $ The average gorilla at the zoo is 19 years old. There is a standard deviation of 6.6 years.